# Homework Help: Find the area enclosed by the parametric equation

1. Feb 26, 2017

### screechy

1. The problem statement, all variables and given/known data
Find the area of the region enclosed by the parametric equation
x = t3- 7t
y = 8t2

2. Relevant equations
A = ∫ y(t) x'(t) dt

3. The attempt at a solution
I initially began with A = ∫ y(t) x'(t) dt
And got to ∫24t4-56t2dt and then to 24∫t4dt - 56∫t2dt
but without a limit/defined area I'm not entirely sure how to proceed.
At first I assumed the question wanted a general equation, in which case I came up with (24/5)t5 - (56/3)t3, but it doesn't seem to be working.

Last edited: Feb 26, 2017
2. Feb 26, 2017

### Eclair_de_XII

It's been a while since I've done this, so someone correct me if I'm wrong. Try factoring $x(t)=t^3-7t$ and $y(t)=8t^2$ and look for the values of $t$ wherein $x=0$ and $y=0$. Those should be your bounds. Again, it's been a while since I've done calculus so someone feel free to correct me.

3. Feb 26, 2017

### screechy

In that case my roots for x would be √7, -√7 and 0, while y would only be 0. What would be my upper and lower bounds?
I just attempted to make them between 0 and √7 but the answer was wrong as well

4. Feb 26, 2017

### LCKurtz

You need to know a bit about what the graph looks like. Apparently it crosses itself, enclosing a loop. If you have a parametric grapher that would be a place to start. Otherwise you can look for a values of $t$ where the graph crosses itself. That would be a values of $t_1$ and $t_2$ where $(x(t_1),y(t_1)) = (x(t_2),y(t_2))$. Use your equations to figure that out. That will give your limits.

5. Feb 27, 2017

### ehild

Eliminate t and find the curve in the (x,y) plane. Yes, it crosses itself at x=0, y=0, which corresponds to t=0. But there is an other crossing point. You get it if you express t as $t=\pm\sqrt{y/8}$ and substitute into the first equation. Because of the plus/minus sign, you get two branches which intersect - where?